1.) velocity: $\vec{v}(t) = < \frac{1}{t}, 2 \cos(2t), -2 \sin(2t) >$, acceleration: $\vec{a}(t) = < -\frac{1}{t^2}, -4 \sin(2t), -4 \cos(2t) >$, speed: $|\vec{v}(t)| = \sqrt{\frac{1}{t^2} + 4}$

2.) Integrate $\vec{a}(t)$ to get $\vec{v}(t)$ and apply the initial condition $\vec{v}(0)$. Same process applied to $\vec{v}(t)$ yields $\vec{r}(t)$.

3.) Start with $\vec{F} = m \vec{a} = <0,-mg>$ where $g = | \vec{a}| = 9.8 m/s^2$. From this it is clear that $\vec{a} = <0, -g>$. Integrate to get $\vec{v}(t) = <0, -gt> + \vec{C}$. But $\vec{C} = \vec{v}(0) = \vec{v_0}$. Integration again (and applying $\vec{r}(0) = 0$) yields $\vec{r}(t) = <0, -\frac{1}{2}gt^2> + t\vec{v_0}$. We write $|\vec{v_0}| = v_0$ (the magnitude of the vector is a scalar), then we can write $$\vec{v_0} = < v_0 \cos \alpha, v_0 \sin \alpha >,$$ where $\alpha$ is the given angle of elevation, $45^o$. Plugging that in yields $\vec{v_0} = < v_0, v_0 >$, but we know that $v_0 = 64$ meters per second, yielding $\vec{v_0} = <64, 64>$. Therefore we have derived $$\vec{r}(t) = < 64t, -\frac{1}{2}gt^2 + 64t >.$$ Thus we have $$\left\{ \begin{array}{ll} x = 64t \\ y = 64t - \frac{1}{2}gt^2 \end{array} \right.$$ range given by solving $y=0$ for $t$ and plugging the solution into $x$, maximum height can be found my maximizing the function defining $y$, and the speed of impact can be found by taking the time found for the range and plugging that into the velocity vector and then taking a magnitude

4.) $\vec{v}(t) = < -\pi \sin(\pi t), \pi \cos(\pi t), 3t^2 >$ and $\vec{a}(t) = < -\pi^2 \cos(\pi t), -\pi^2 \sin(\pi t), 6t >$, so use $a_T = \frac{\vec{v} \cdot \vec{a}}{|\vec{v}|}$ and $a_N = \frac{|\vec{v} \times \vec{a}|}{|\vec{v}|^3}$

5.) $\vec{v}(t) = < 2t, 1, 6t - 17 >$ so $|\vec{v}(t)| = \sqrt{4t^2 + 1 + 36t^2 - 102t + 289}$. Minimize this function.

6.) Recall that $\vec{F} = m\vec{a}$. So find $\vec{a} = < 18t, 14, 18t>$ and then compute $\vec{F}$.

7.)

8.) Suppose that $\vec{v}(t) = < f(t), g(t), h(t) >$. Then $\vec{a}(t) = < f'(t), g'(t), h'(t) >$. So we want to show that the dot product of velocity and accleration is zero, that is $$\vec{v}(t) \cdot \vec{a}(t) = f(t)f'(t) + g(t)g'(t) + h(t)h'(t) = 0.$$ So, since the speed is constnat we have $$C = |\vec{v}(t)| = \sqrt{f(t)^2 + g(t)^2 + h(t)^2}$$ for some constant $C$. This is equivalent to $$C^2 = f(t)^2 + g(t)^2 + h(t)^2.$$ Taking a derivative yields $$0 = 2f(t)f'(t) + 2g(t)g'(t) + 2h(t)h'(t).$$ Division by $2$ will then yield the result.

9a.) Domain is all of $\mathbb{R}^2$.

9b.) Domain is all of $\mathbb{R}^2$ except for the $y$-axis.

9c.) Domain is all of $\mathbb{R}^2$ except the $y$-axis and the $x$-axis.

10.) For a general $k$ you get $k = x^3 - y$ also known as $y = x^3 - k$, which are shifts of a simple cubic function.